# indefinite integral

#### paulmdrdo

##### Active member
how would i go about solving this

$\displaystyle\int\frac{\arctan x}{1+x^2}dx$?

i tried substitution but i didn't work.

Last edited:

#### HallsofIvy

##### Well-known member
MHB Math Helper
That's very strange. Did you try the obvious substitution $$u= tan^{-1}(x)$$. What is the derivative of $$tan^{1}(x)$$?

(And do you understand that $$tan^{-1}(x)$$ here is the arctangent, NOT the $$\frac{1}{tan(x)}$$?!)

#### paulmdrdo

##### Active member
yes, i know that. but how do i use substitution here?

#### Ackbach

##### Indicium Physicus
Staff member
Following HoI's suggestion, we get that $u=\tan^{-1}(x)$, and
$$du= \frac{1}{1+x^{2}} \, dx,$$
and the integral becomes
$$\int u \, du.$$
Can you continue?

#### paulmdrdo

##### Active member
this is my answer,

$\displaystyle\frac{1}{2}(\tan^{-1}x)^2+C$

Last edited:

#### Plato

##### Well-known member
MHB Math Helper
this is my answer,

$\displaystyle\frac{1}{2}(Tan^{-1})^2+C$
That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$

#### topsquark

##### Well-known member
MHB Math Helper
That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$
I find that "atn(x)" works quite well also.

-Dan